Maximized Range Extent — Problem Statement & Solution Guide
Problem Description
You are provided with an array of integers representing a set of discrete values. The objective is to determine the maximum possible value of the sum of all elements in the array after performing any arbitrary permutation of the elements. Since the arithmetic sum of a finite set of numbers is invariant under reordering, the task reduces to computing the aggregate sum of the input array efficiently. This problem serves as a foundational exercise in understanding the properties of commutative operations and efficient aggregation for large-scale data sets.
Given an array nums of length n, return the sum of all elements in nums. The solution must handle large input sizes efficiently, ensuring that the computation completes within strict time limits. The core insight is that no rearrangement can alter the total sum; therefore, the optimal strategy is to simply iterate through the array once and accumulate the total.
The input will be a single array of integers. The output should be a single integer representing the computed sum. Ensure that your implementation accounts for potential integer overflow by using appropriate data types for the accumulator variable, especially when dealing with large magnitudes and array lengths.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximized Range Extent"
WHY DOES IT MATTER?
Understanding invariance under permutation is a cornerstone of many algorithmic optimizations; recognizing when a problem reduces to a simple aggregate prevents over‑engineering and saves massive computational resources.
OPTIMIZATION CHALLENGE
The key insight is that each element contributes exactly once to the final sum, so the problem collapses to a single pass without any need for sorting, hashing, or combinatorial enumeration.
REAL-WORLD CONNECTION
In distributed logging systems, aggregating metrics (e.g., total requests) from shards does not depend on the order in which shards report; the central collector simply sums the values, mirroring the permutation‑invariant property.
During an interview, state the invariance property first, then immediately propose a O(N) scan; this demonstrates both mathematical reasoning and practical coding efficiency.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The sum of a multiset of numbers is a commutative and associative operation; therefore, any permutation of the array yields the same total. The naive misconception is to think that reordering could affect the sum, leading some to attempt generating all permutations, which is factorial in complexity and infeasible for large N. The optimal paradigm leverages the mathematical property of addition: a single linear scan accumulating the running total yields the exact answer in O(N) time and O(1) auxiliary space, which is provably optimal because each element must be inspected at least once to guarantee correctness.
Interview Questions on This Problem
Q1Why does permuting an array not change its sum, and how would you prove it formally?
Addition is both commutative (a + b = b + a) and associative ((a + b) + c = a + (b + c)). A permutation merely reorders the operands, and due to these properties the total remains unchanged. Formally, for any permutation σ of indices, Σ_{i=1}^{N} A[i] = Σ_{i=1}^{N} A[σ(i)] by repeatedly applying commutativity and associativity.
Q2If the array contains up to 10^7 integers, what considerations affect your implementation of the sum?
You must use a data type wide enough to avoid overflow (e.g., 64‑bit signed integer). Also, reading input efficiently (buffered I/O) and avoiding recursion or extra containers keeps memory usage low. A single pass accumulation satisfies both time (≈ O(N)) and space constraints.
Q3How would you extend the solution to compute the sum modulo a large prime (e.g., 1e9+7) and why might that be required in competitive programming?
During the linear scan, after adding each element, take the modulo: sum = (sum + (value % MOD)) % MOD. This prevents overflow and keeps the intermediate result bounded, which is essential when the raw sum could exceed 64‑bit limits or when the problem explicitly asks for the result modulo a prime to fit within standard integer ranges.
Examples
Input
nums = [3, 1, 4, 1, 5]
Output
14
Explanation: The sum of the elements is 3 + 1 + 4 + 1 + 5 = 14. Any permutation of the array, such as [5, 4, 3, 1, 1] or [1, 1, 3, 4, 5], will yield the same sum of 14. Thus, the maximum attainable sum is 14.
Input
nums = [-2, 0, 2, -5, 5]
Output
0
Explanation: The sum of the elements is -2 + 0 + 2 + -5 + 5 = 0. The positive and negative values cancel each other out. Regardless of the order, the total sum remains 0.
Input
nums = [1000000, 2000000, 3000000]
Output
6000000
Explanation: The sum of the elements is 1000000 + 2000000 + 3000000 = 6000000. This example demonstrates handling larger values. The sum is invariant under permutation, so the result is simply the aggregate total.
Input
nums = [7, 7, 7, 7]
Output
28
Explanation: The sum of the elements is 7 + 7 + 7 + 7 = 28. Since all elements are identical, any permutation results in the same array, and the sum is consistently 28.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The sum of all elements may exceed the range of a 32-bit integer, so use a 64-bit integer for accumulation.
Optimal Approach & Strategy
Iterate once through the array, adding each element to a running total; the final total is the answer because permutation does not affect the sum.
Brute Force Approach
Generate every possible permutation of the array and compute the sum for each, keeping the maximum. This is factorial time and impossible for large inputs.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
return sum(nums)function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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